229 is another term because 613^2 divides 229!+1. See A115091 for primes whose square divides m!+1 for some m. An examination of the factorizations of m!+1 for m<=100 found no additional squares.  T. D. Noe, Mar 01 2006
562 is also a term because 562!+1 is divisible by 563^2.  Vladeta Jovovic, Mar 30 2004
Comment from Francois BRUNAULT, Nov 23 2008: A web search reveals that for 1 <= k <= 228 there are 82 values of k for which k! + 1 has not been completely factored (the smallest is k=103), so showing that 229 and 562 are indeed the next two terms will be a huge task. I checked that k!+1 is not divisible by p^2 for k <= 1000 and prime p < 10^8.
It is very likely that 229 and 562 are the next two terms, but this has not yet been proved.  Nov 29 2008
Contains A007540(n)1 for all n. That sequence is conjectured to be infinite.  Robert Israel, Jul 04 2016
This sequence includes A146968 (solutions of Brocard's problem).  Salvador CerdÃ¡, Mar 08 2016
I may be wrong, but perhaps it is not yet proved that there are no terms between 23 and 228?  N. J. A. Sloane, Jul 24 2017
If k > 562 and k! + 1 is divisible by p^2 where p is prime, then either k > 10000 or p > 2038074743 (the hundred millionth prime).  Jason Zimba, Oct 21 2021
